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If \((l, m)\) is the circumcentre of an equilateral triangle inscribed in the ellipse…
If \((l, m)\) is the circumcentre of an equilateral triangle inscribed in the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) having vertices at points with eccentric angles \(\theta_1, \theta_2\) and \(\theta_3\), then \(\frac{2}{3}\left[\cos \left(\theta_1-\theta_2\right)+\cos \left(\theta_2-\theta_3\right)+\cos \left(\theta_3-\theta_1\right)\right]=\)
\(\frac{9 l^2}{2 a^2}+\frac{9 m^2}{b^2}-1\) \(\frac{l^2}{\mathrm{a}^2}+\frac{m^2}{b^2}-3\) \(\frac{3 l^2}{a^2}+\frac{3 m^2}{b^2}-1\) \(\frac{3 l^2}{a^2}+\frac{3 m^2}{b^2}-\frac{3}{2}\) Solution
As we know coordinates of circumcentre of an equilateral triangle is same as centroid, so
\(\begin{aligned}
& (l, m)=\left(\frac{a \cos \theta_1+a \cos \theta_2+a \cos \theta_3}{3},\frac{b \sin \theta_1+b \sin \theta_2+b \sin \theta_3}{3}\right) \\
& \Rightarrow \quad \frac{3 l}{a}=\cos \theta_1+\cos \theta_2+\cos \theta_3 \quad \ldots \text { (i) } \\
& \text { and } \quad \frac{3 m}{b}=\sin \theta_1+\sin \theta_2+\sin \theta_3 \quad \ldots \text { (ii) }
\end{aligned}\)
Now, after squaring and adding the Eqs. (i) and (ii), we get
\(\begin{aligned}
& \frac{9 l^2}{a^2}+\frac{9 m^2}{b^2}=\Sigma \cos ^2 \theta_1+2 \Sigma \cos \theta_1 \cos \theta_2+\Sigma \sin ^2 \theta_1 \\
& \Rightarrow \frac{9 l^2}{a^2}+\frac{9 m^2}{b^2}=3+2 \Sigma \sin \theta_1 \sin \theta_2 \\
& \Rightarrow \frac{2}{3} \Sigma \cos \left(\theta_1-\theta_2\right)=\frac{3 l^2}{a^2}+\frac{3 m^2}{b^2}-1 \\
& \Rightarrow \frac{2}{3}\left[\cos \left(\theta_1-\theta_2\right)+\cos \left(\theta_2-\theta_3\right)+\cos \left(\theta_3-\theta_1\right)\right] \\
& =\frac{3 l^2}{a^2}+\frac{3 m^2}{b^2}-1
\end{aligned}\)
Asked in: AP EAMCET 2019 (22 Apr Shift 1)
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